Problem 29 - Chemistry: The Central Science

Question

Molybdenum metal must absorb radiation with a minimum frequency of \(1.09 \times 10^{15} \mathrm{~s}^{-1}\) before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(120 \mathrm{~nm}\), what is the maximum possible kinetic energy of the emitted electrons?

Step-by-Step Solution

Answer

(a) The minimum energy needed to eject an electron is \(7.22 \times 10^{-19} J\). (b) The wavelength of radiation needed to provide a photon of this energy is \(275 nm\). (c) The maximum possible kinetic energy of the emitted electrons is \(9.38 \times 10^{-19} J\).

Step 1: Formula to calculate the energy of a photon

In this part, we need to calculate the energy of a photon with the given frequency. We use the equation \(E = hf\).

Step 2: Plug in the given frequency and calculate the energy

Given frequency is \(1.09 \times 10^{15} s^{-1}\). Therefore, the minimum energy needed to eject an electron is: \(E = (6.626 \times 10^{-34} J \cdot s)(1.09 \times 10^{15} s^{-1}) = 7.22 \times 10^{-19} J\). So the minimum energy needed to eject an electron is \(7.22 \times 10^{-19} J\). #b) Finding the wavelength of radiation to provide the minimum photon energy#

Step 3: Use the wavelength-frequency relation to find the wavelength

We know the frequency of the photon, and we can use the relation \(f = \frac{c}{\lambda}\) to find the wavelength. Rearranging the formula, we get: \(\lambda = \frac{c}{f}\)

Step 4: Plug in the values and calculate the wavelength

Now we can plug in the given frequency and speed of light to find the wavelength: \(\lambda = \frac{3.00 \times 10^8 m/s}{1.09 \times 10^{15} s^{-1}} = 2.75 \times 10^{-7} m = 275 nm\) The wavelength of radiation needed to provide a photon of this energy is \(275 nm\). #c) Finding the maximum possible kinetic energy of the emitted electrons#

Step 5: Calculate the energy of the photon with a wavelength of 120 nm

First, we need to find the frequency of light with a wavelength of \(120 nm\). Using the relation \(f = \frac{c}{\lambda}\), we can find the frequency: \(f = \frac{3.00 \times 10^8 m/s}{120 \times 10^{-9} m} = 2.50 \times 10^{15} s^{-1}\) Now we can find the energy of the photon using \(E = hf\): \(E = (6.626 \times 10^{-34} J \cdot s)(2.50 \times 10^{15} s^{-1}) = 1.66 \times 10^{-18} J\)

Step 6: Calculate the maximum kinetic energy of the emitted electrons

Using the equation \(K_{max} = hf - W_0\), we can find the maximum kinetic energy of the emitted electrons: \(K_{max} = (1.66 \times 10^{-18} J) - (7.22 \times 10^{-19} J) = 9.38 \times 10^{-19} J\) The maximum possible kinetic energy of the emitted electrons is \(9.38 \times 10^{-19} J\).

Key Concepts

Molybdenum

Molybdenum is a chemical element known for its strong and durable properties. In the context of the photoelectric effect, molybdenum exhibits interesting behavior due to its relatively high work function. The work function is the minimum energy required to remove an electron from the surface of a material. For molybdenum, this work function corresponds to a minimum frequency of radiation of \(1.09 \times 10^{15} \, \text{s}^{-1}\) necessary to eject electrons.

• Molybdenum's properties make it valuable in scientific experiments, where accurate and repeatable material response is needed.
• The high work function implies that molybdenum requires photons with considerable energy to initiate the emission of electrons.
• This property is key in studies exploring the photoelectric effect, providing insight into quantum mechanics and electron behavior.

Photon Energy

Photon energy is a crucial concept in understanding the photoelectric effect. It refers to the energy carried by a photon, which is the basic unit of light and electromagnetic radiation. In the context of the photoelectric effect involving molybdenum, as seen in the problem, the equation \(E = hf\) is applied to calculate photon energy, where \(E\) is the energy, \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\), and \(f\) is the frequency of the photon.

• Photon energy must equal or exceed the work function to eject an electron from a surface.
• This concept also shows the quantized nature of light, suggesting that light can be absorbed only in discrete quanta.
• For molybdenum, a photon energy of \(7.22 \times 10^{-19} \, \text{J}\) is needed to remove an electron from the metal surface.

Wavelength Calculation

Understanding the relationship between the frequency and wavelength of light is essential for calculating the specific wavelength needed for certain photon energies. The speed of light \(c\) links these two properties through the formula \(f = \frac{c}{\lambda}\), where \(f\) is the frequency, \(\lambda\) is the wavelength, and \(c\) is the speed of light \(3.00 \times 10^8 \, \text{m/s}\).

• Rearranging the formula allows us to calculate the wavelength as \(\lambda = \frac{c}{f}\).
• For molybdenum, with a minimum frequency of \(1.09 \times 10^{15} \, \text{s}^{-1}\), the corresponding wavelength of light is \(275 \, \text{nm}\).
• This helps in determining what wavelength of radiation must be used in experiments to ensure electrons are emitted from specific materials.

Kinetic Energy of Electrons

The kinetic energy of electrons emitted in the photoelectric effect is a key aspect of understanding electron behavior. Once a photon provides enough energy to overcome the work function of a material, any additional energy is converted into the kinetic energy of the ejected electron. This is given by the equation \(K_{\text{max}} = hf - W_0\) where \(W_0\) is the work function.

• The kinetic energy depends on both the frequency of the incoming light and the work function of the material.
• In the molybdenum example, when irradiated with a wavelength of \(120 \, \text{nm}\), the emitted electrons possess a maximum kinetic energy of \(9.38 \times 10^{-19} \, \text{J}\).
• This illustrates how different wavelengths of light can affect the kinetic energy of emitted electrons, and thus the insights gathered from such experiments inform applications in solar cells and photodetectors.